Question

A cell of e.m.f. $E$ is connected to a resistance $R_1$ for time $t$ and the amount of heat generated in it is $H$. If the resistance $R_1$ is replaced by another resistance $R_2$ and is connected to the cell for the same time $t$, the amount of heat generated in $R_2$ is $4H$. Then the internal resistance of the cell is

• A. $\frac{2R_{1}+R_{2}}{2}$
• B. $\sqrt{R_{1}R_{2}} \frac{2\sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2\sqrt{R_{1}}}$
• C. $\sqrt{R_{1}R_{2}} \frac{\sqrt{R_{2}}-2\sqrt{R_{1}}}{2\sqrt{R_{2}} -\sqrt{R_{1}}}$
• D. $\sqrt{R_{1}R_{2}} \frac{\sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-\sqrt{R_{1}}}$

Answer 1

Answer: (B)

We have, $H=l^{2} R$
According to given condition, $l_{1}^{2} R_{1}=H$
and $ l_{2}^{2} R_{2}=4 H$
$\Rightarrow \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H$
and $\frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H$
$\therefore \frac{R_{2}}{\left(R_{2}+r\right)^{2}}=\frac{4 R_{1}}{\left(R_{1}+r\right)^{2}}$
$\Rightarrow \sqrt{R_{2}}\left(R_{1}+r\right)=2 \sqrt{R_{1}}\left(R_{2}+r\right)$
On solving, $r=\frac{\sqrt{R_{1} R_{2}}\left[\sqrt{R_{1}}-2 \sqrt{R_{2}}\right]}{2 \sqrt{R_{1}}-\sqrt{R_{2}}}$